
In the figure, \[{S_1}\]and \[{S_2}\]are identical springs. The oscillation frequency of the mass m is\[f\]. If one spring is removed, the frequency will become

Fig: The springs attached to the mass m
A. \[f\]
B. \[f \times 2\]
C. \[f \times \sqrt 2 \]
D. \[\dfrac{f}{{\sqrt 2 }}\]
Answer
163.2k+ views
Hint:We use the frequency of the spring-block system by finding the equivalent of the springs connected to the block. The frequency of the spring-block system is proportional to the square root of the spring constant.
Formula used:
\[\dfrac{1}{{{k_{eq}}}} = \dfrac{1}{{{k_1}}} + \dfrac{1}{{{k_2}}} + \dfrac{1}{{{k_3}}} \ldots + \dfrac{1}{{{k_n}}}\], here \[{k_{eq}}\]is the equivalent spring constant when the springs are connected in series.
\[{k_{eq}} = {k_1} + {k_2} + {k_3} \ldots + {k_n}\], here \[{k_{eq}}\] is the equivalent spring constant when the springs are connected in parallel.
\[f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{{K_{eq}}}}{m}} \],
where f is the frequency of spring-mass system.
Complete step by step solution:

Fig: The springs attached to the mass m
As the springs are identical, so the spring constant of both springs will be equal.
Let the spring constant of the given two springs is k.
Both springs are connected in series. So, the equivalent spring constant is,
\[{k_{eq}} = k + k\]
\[{k_{eq}} = 2k\]
So, the frequency of the initial spring-block system is,
\[f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{2k}}{m}} \]
When one of the spring is removed then the net spring constant of the system will be k,
So, the frequency of the final spring-block system is,
\[{f_2} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} \]
On dividing both the equations, we get
\[\dfrac{f}{{{f_2}}} = \dfrac{{\dfrac{1}{{2\pi }}\sqrt {\dfrac{{2k}}{m}} }}{{\dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} }}\]
\[\dfrac{f}{{{f_2}}} = \sqrt {\dfrac{{2k}}{k}} \]
\[{f_2} = \dfrac{f}{{\sqrt 2 }}\]
So, the frequency of the spring-block system when one of the spring is removed will be \[\dfrac{f}{{\sqrt 2 }}\]
Therefore, the correct option is (D).
Note:We should be careful while finding the equivalent spring constant of the combination of the springs. When the compression in all the springs are same when block is displaced, then the springs are in parallel combination.
Formula used:
\[\dfrac{1}{{{k_{eq}}}} = \dfrac{1}{{{k_1}}} + \dfrac{1}{{{k_2}}} + \dfrac{1}{{{k_3}}} \ldots + \dfrac{1}{{{k_n}}}\], here \[{k_{eq}}\]is the equivalent spring constant when the springs are connected in series.
\[{k_{eq}} = {k_1} + {k_2} + {k_3} \ldots + {k_n}\], here \[{k_{eq}}\] is the equivalent spring constant when the springs are connected in parallel.
\[f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{{K_{eq}}}}{m}} \],
where f is the frequency of spring-mass system.
Complete step by step solution:

Fig: The springs attached to the mass m
As the springs are identical, so the spring constant of both springs will be equal.
Let the spring constant of the given two springs is k.
Both springs are connected in series. So, the equivalent spring constant is,
\[{k_{eq}} = k + k\]
\[{k_{eq}} = 2k\]
So, the frequency of the initial spring-block system is,
\[f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{2k}}{m}} \]
When one of the spring is removed then the net spring constant of the system will be k,
So, the frequency of the final spring-block system is,
\[{f_2} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} \]
On dividing both the equations, we get
\[\dfrac{f}{{{f_2}}} = \dfrac{{\dfrac{1}{{2\pi }}\sqrt {\dfrac{{2k}}{m}} }}{{\dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} }}\]
\[\dfrac{f}{{{f_2}}} = \sqrt {\dfrac{{2k}}{k}} \]
\[{f_2} = \dfrac{f}{{\sqrt 2 }}\]
So, the frequency of the spring-block system when one of the spring is removed will be \[\dfrac{f}{{\sqrt 2 }}\]
Therefore, the correct option is (D).
Note:We should be careful while finding the equivalent spring constant of the combination of the springs. When the compression in all the springs are same when block is displaced, then the springs are in parallel combination.
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